We use the variational procedure used in [4] with a few modifications prompted by the presence of the functionh

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

ASYMPTOTIC PROFILE OF A RADIALLY SYMMETRIC SOLUTION WITH TRANSITION LAYERS FOR AN

UNBALANCED BISTABLE EQUATION

HIROSHI MATSUZAWA

Abstract. In this article, we consider the semilinear elliptic problem

−ε²∆u=h(|x|)²(u−a(|x|))(1−u²)

in B1(0) with the Neumann boundary condition. The function a is a C¹ function satisfying |a(x)| < 1 for x ∈ [0,1] and a⁰(0) = 0. In particular we consider the casea(r) = 0 on some intervalI⊂[0,1]. The functionhis a positiveC¹function satisfyingh⁰(0) = 0. We investigate an asymptotic profile of the global minimizer corresponding to the energy functional asε→0. We use the variational procedure used in [4] with a few modifications prompted by the presence of the functionh.

1. Introduction and Statement of Main Results In this article, we consider the boundary value problem

−ε²∆u=h(|x|)²(u−a(|x|))(1−u²) in B₁(0)

∂u

∂ν = 0 on∂B1(0)

(1.1) where ε is a small positive parameter,B1(0) is a unit ball in R^N centered at the origin, and the functionais aC¹function on [0,1] satisfying−1< a(|x|)<1 and a⁰(0) = 0. The function his a positive C¹ function on [0,1] satisfying h⁰(0) = 0.

We setr=|x|.

Problem (1.1) appears in various models such as population genetics, chemical reactor theory and phase transition phenomena. See [1] and the references therein.

If the function h satisfies h(r) ≡ 1 and the function a satisfies a(r) 6≡ 0, then this problem (1.1) has been studied in [1], [4] and [7]. In this case, it is shown that there exist radially symmetric solutions with transition layers near the set {x∈B₁(0)|a(|x|) = 0}. If the set{r∈R|a(r) = 0} contains an intervalI, then the problem to decide the configuration of transition layer onI is more delicate.

When N = 1, if the functionh satisfies h(r) 6≡ 1 and the function a satisfies a(r) ≡ 0, then problem (1.1) has been studied in [8] and [9]. In this case, it is

2000Mathematics Subject Classification. 35B40, 35J25, 35J55, 35J50, 35K57.

Key words and phrases. Transition layer; Allen-Cahn equation; bistable equation; unbalanced.

c

2006 Texas State University - San Marcos.

Submitted August 31, 2005. Published January 11, 2006.

1

shown that there exist stable solutions with transition layers near prescribed local minimum points ofh.

In this paper, we consider the case where the functionasatisfies a(r)6≡0 with a(r) = 0 on some intervalI⊂(0,1). We show the minimum point of the function r^N−1h(r) on I has very important role to decide the configuration of transition layer onI in this case.

We note that in [4], Dancer and Shusen Yan considered a problem similar to ours. They assume thatN ≥2,h≡1 and the nonlinear term isu(u−a|x|)(1−u) satisfying a(r) = 1/2 on I = [l1, l2] and a(r) < 1/2 for l1−r > 0 small and a(r) > 1/2 for r−l2 > 0 small, then a global minimizer of the corresponding functional has a transition layer near the l1, that is, the minimum point of r^N−1 onI (see [4, Theorem 1.3]). In this sense, we can say that our results are natural extension of the results in [4]. We are going to follow throughout the variational procedure used in [4] with a few modifications prompted by the presence of the functionh.

Here we state the energy functional, corresponding to (1.1), Jε(u) =

Z

B₁(0)

ε²

2 |∇u|²−F(|x|, u)dx, where F(|x|, u) =Ru

−1f(|x|, s)dsand f(|x|, u) =h(|x|)²(u−a(|x|))(1−u²). It is easy to see that the following minimization problem has a minimizer

inf{Jε(u)|u∈H¹(B1(0))}. (1.2) LetA₋={x∈B1(0)|a(|x|)<0} andA+={x∈B1(0)|a(|x|)>0}.

In this paper, we will analyze the profile of the minimizer of (1.2), and prove the following results.

Theorem 1.1. Letuεbe a global minimizer of (1.2). Thenuεis radially symmetric and

uε→

(1, uniformly on each compact subset of A₋,

−1, uniformly on each compact subset of A+,

as ε→ 0. In particular u_ε converges uniformly near the boundary of B₁(0), that is, if a(r) < 0 on [r0,1] for some r0 > 0, uε → 1 uniformly on B1(0)\Br₀(0) and if a(r)>0 on [r₀,1] for some r₀ >0,u_ε → −1 uniformly on B₁(0)\B_r0(0).

Moreover, for any0< r1≤r2<1 with a(ri) = 0,i= 1,2,a(r)6= 0 forr1−r >0 small and for r−r2>0 small, a(r) = 0 ifr∈[r1, r2], we have:

(i) If a(r) < 0 for r₁−r > 0 small and a(r) > 0 for r−r₂ > 0, then for any smallη >0 and for any smallθ >0, there exists a positive numberε0

which has the following properties:

(a) For allε∈(0, ε0], there existtε,1< tε,2 such that u_ε(r)>1−η forr∈[r₁−θ, t_ε,1),

uε(tε,1) = 1−η, uε(tε,2) =−1 +η,

u_ε(r)<−1 +η, forr∈(t_ε,2, r₂+θ].

(b) The functionuε(r)is decreasing on the interval(tε,1, tε,2)

(c) The inequality0 < R₁ ≤ ^tε,2−t_ε ^ε,1 ≤R₂ holds, where R₁ and R₂ are two constants independent ofε >0.

(d) Iftε_j,1,tε_j,2→tfor some positive sequence{εj}converging to zero as j→ ∞, thent satisfiesh(t)t^N−1= min_s∈[r1,r2]h(s)s^N−1.

(ii) If a(r)>0forr1−r >0small and a(r)<0 forr−r2>0, then for each smallη >0and for each smallθ >0, there exists a positive numberε0which has the following properties: For eachε∈(0, ε0], there existtε,1< tε,2such that

(a)

u_ε(r)<−1 +η forr∈[r₁−θ, t_ε,1), uε(tε,1) =−1 +η,

uε(tε,2) = 1−η,

u_ε(r)>1−η, forr∈(t_ε,2, r₂+θ].

(b) The functionuε(r)is increasing in (tε,1, tε,2).

(c) The inequality0 < R₁ ≤ ^tε,2−t_ε ^ε,1 ≤R₂ holds, where R₁ and R₂ are two constants independent ofε >0.

(d) Ift_εj,1,t_εj,2→tfor some positive sequence{ε_j}converging to zero as j→ ∞, thent satisfiesh(t)t^N−1= min_s∈[r1,r₂]h(s)s^N−1.

Figure 1. Profile of the global minimizeru_ε

Remarks.

• Note that results from (a) to (c) both in cases (i) and (ii) are not related to the presence of the functionh. The effect of presence of functionhappears in the result (d) in (i) and (ii).

• If min_s∈[r1,r2]s^N−1h(s) is attained at a unique point t, we can show tε,1, tε,2→tasε→0 without taking subsequences.

• If the functionr^N−1h(r) is constant on [r1, r2], it is a very difficult problem to know the location of the pointt∈[r1, r2].

This paper is organized as follows: In section 2, we present some preliminary results. In section 3, we prove the main theorem.

2. Preliminary Results

LetD is a bounded domain in R^N. Letf(x, t) be a function defined onD×R which is bounded onD×[−1,1]. Supposef is continuous ont∈Rfor eachx∈D

and is measurable inDfor eacht∈R. We also assume f(x, t)>0 forx∈D, t <−1;

f(x, t)<0 forx∈D, t >1. (2.1) Consider the minimization problem

inf

Jε(u, D) :=

Z

D

ε²

2|∇u|²−F(x, u)dx:u−η∈H₀¹(D)

, (2.2)

whereη∈H¹(D) with−1≤η≤1 onD and F(x, t) =

Z t

−1

f(x, s)ds.

We can prove next two lemmas by methods similar to [4]. For the readers conve- nience, we prove these lemmas in this section.

Lemma 2.1. Suppose that f(x, t) satisfies (2.1). Let u_ε be a minimizer of (2.2).

Then−1≤u_ε≤1 onD.

Proof. We prove−1≤u_ε onD. LetM ={x:u_ε(x)<−1}. Define ˜u_εby

˜ uε(x) =

(uε(x) ifx∈D\M

−1 ifx∈M.

Since uε(x) =η ≥ −1 on∂D, we see thatM is compactly contained inD. Thus

˜

u−η ∈ H₀¹(D). If the measure m(M) of M is positive, we have J_ε(˜u_ε, D) <

Jε(uε, D). Because uε is a minimizer, we see m(M) = 0, where m(A) denotes the Lebesgue measure of the set A. Thus uε ≥ −1. Similarly we can prove that

uε≤1.

Lemma 2.2. Suppose that f₁(x, t) and f₂(x, t) both satisfy (2.1) and the same regularity assumption on f. Assume that η_i ∈ H¹(D) satisfy −1 ≤η_i ≤1 on D for i = 1,2. Let uε,i be a corresponding minimizer of (2.2), where f = f_i and η =η_i, i = 1,2. Suppose that f₁(x, t) ≥f₂(x, t) for all (x, t)∈ D×[−1,1] and 1≥η₁≥η₂≥ −1. Thenu_ε,1≥u_ε,2.

Proof. LetM ={x∈D:u_ε,2> u_ε,1}. Define ϕ_ε= (u_ε,2−u_ε,1)⁺. Sinceη₁≥η₂, we haveϕ_ε∈H₀¹(D). SetF_i(x, u) =Ru

−1f_i(x, s)ds. Sinceu_ε,i is a minimizer of Jε,i(u) :=

Z

D

ε²

2|∇u|²−Fi(x, u)dx andϕε= 0 forx∈D\M, we have

0≤Jε,1(uε,1+ϕε)−Jε,1(uε,1)

= Z

M

ε²

2 (|∇(uε,1+ϕε)|²− |∇uε,1|²)dx− Z

M

Z u_ε,1+ϕ_ε uε,1

f₁(x, s)ds

≤ Z

M

ε²

2 (|∇(uε,1+ϕε)|²− |∇uε,1|²)dx− Z

M

Z u_ε,1+ϕ_ε u_ε,1

f₂(x, s)ds

=J_ε,2(u_ε,2)−J_ε,2(u_ε,2−ϕ_ε)≤0.

This implies thatuε,1+ϕεis also a minimizer ofJε,1(u). LetL >0 be large enough such thatf₁(x, t) +Ltis strictly increasing forx∈D,t∈[−1,1]. From

−ε²∆(uε,1+ϕε) =f₁(uε,1+ϕε), we obtain

−ε²∆ϕ_ε=f₁(u_ε,1+ϕ_ε)−f₁(u_ε,1).

Thus

−ε²∆ϕε+Lϕε=f₁(uε,1+ϕε) +L(uε,1+ϕε)−(f₁(uε,1) +Luε,1)>0 in D. Fix z₀ ∈M. Let x₀ ∈∂M such that |x₀−z₀|= dist(z₀, ∂M). Using the Strong maximum principle and Hopf’s lemma in B_{dist(z0,∂M)}(z₀), we obtain that

∂ϕ_ε

∂ν (x0) < 0, where ν = (x0−z0)/|x0−z0|. But ϕε(x) = 0 for x /∈ M. Thus,

∂ϕ_ε

∂ν (x0) = 0. This is a contradiction. Thus we obtainM =∅.

3. Proof of Main Theorem

To prove Theorem 1.1, the following proposition is used as the first step.

Propositon 3.1. Let u_ε be a global minimizer of the problem (1.2). Then u_ε satisfies

uε→

(1 uniformly on each compact subset of A−

−1 uniformly on each compact subset of A₊ asε→0.

Proof. Let x0 ∈ A₋. Choose δ > 0 small so that Bδ(x0) ⊂⊂ A. Take b ∈ (max_z∈B

δ(x₀)a(z),1/2). Define fx₀,δ,b(t) = (min_z∈Bδ(x0)h(|z|)²)(t −b)(1−t²).

Then for x ∈ Bδ(x0), t ∈ [−1,1], we have f(|x|, t) ≥ fx₀,δ,b(t). Let uε,x₀,δ,b be the minimizer of

infnZ

B_δ(x₀)

ε²

2 |∇u|²−Fx₀,δ,b(u)dx:u+ 1∈H₀¹(Bδ(x0))o ,

whereFx₀,δ,b(t) =Rt

−1fx₀,δ,b(s)ds. It follows from Lemmas 2.1 and 2.2 that u_ε,x0,δ,b(x)≤u_ε(x)≤1, forx∈B_δ(x₀).

Since R1

−1fx₀,δ,b(s)ds > 0, it follows from [2, 3] that uε,x₀,δ,b(x) → 1 as ε → 0 uniformly inB_δ/2(x0), thusuε(x)→1 asε→0 uniformly inB_δ/2(x0).

To prove the rest of Theorem 1.1, we need the following proposition and lemma.

Propositon 3.2. Let ube a local minimizer of the problem infnZ

B₁(0)

1

2|∇u|²−G(|x|, u)dx:u∈H¹(B₁(0))o .

HereG(r, t) =Rt

−1g(r, s)ds,g(r, t)isC¹int∈Rfor eachr≥0,g(r, t)andg_t(r, t) are measurable on [0,+∞) for each t ∈ R, g(r, t) < 0 if t < −1 or t > 1 and

|g(r, t)|+|gt(r, t)| is bounded on [0, k]×[−2,2] for any k >0. Then u is radial, i.e., u(x) =u(|x|).

The proof of the above proposition can be found in [4, Proposition 2.6].

Lemma 3.3. Let 0< η <1 be any fixed constant andwsatisfies

−wzz=w(1−w²) on R, w(0) =−1 +η (resp. w(0) = 1−η), w(z)≤ −1 +η (resp. w(z)≥1−η) forz≤0,

w is bounded onR. Thenwis a unique solution of

−w_zz=w(1−w²) on R, w(0) =−1 +η (resp. w(0) = 1−η), w⁰(z)>0 (resp. w⁰(z)<0) z∈R, w(z)→ ±1 (resp. w(z)→ ∓1) asz→ ±∞.

The proof of the above lemma can be found in [6]. Now we prove the rest of Theorem 1.1.

Proof of Theorem 1.1. For the sake of simplicity, we prove for the case wherea(r)<

0 on [0, r1), a(r) = 0 on [r1, r2] anda(r)>0 on (r2,1] for some 0 < r1 < r2 <1 (see Figure 1 in Section 1).

Part 1. First we show that uε converges uniformly near the boundary of B1(0), that is,uε→ −1 uniformly onB1(0)\Br₂+τ(0) for any small τ >0. We note that we have u_ε → −1 uniformly onB_1−τ(0)\Br2+τ(0) as ε→ 0. Now we claim that u_ε(r)≤u_ε(1−τ) =:T_εforr∈[1−τ,1]. We define the function ˜u_εby

˜ u_ε(r) =







u_ε(r) ifr∈[0,1−τ]

uε(r) ifuε(r)< Tε andr∈[1−τ,1], Tε ifuε(r)≥Tε andr∈[1−τ,1].

We note that ˜uε∈H¹(B1(0)) and−F(r, Tε)≤ −F(r, t) forε >0 and |r−1|small and t ≥T_ε. Hence we obtain J_ε(˜u_ε)< J_ε(u_ε) and we have a contradiction if we assume that the measure of the set {r ∈ [0,1]|uε(r) > T_ε and r ∈ [1−τ,1]} is positive. Hence−1< uε(r)≤Tεanduε→ −1 uniformly on B1(0)\Br₂+τ(0).

Part 2. We remark that, by Proposition 3.1,uεis radially symmetric and we note that for anyt2> t1,uε is a minimizer of the following problem

inf{J_ε(u, B_t2(0)\B_t1(0)) :u−u_ε∈H₀¹(B_t2(0)\B_t1(0))}, where

J_ε(u, M) = Z

M

ε²

2|∇u|²−F(|x|, u)dx

for any open setM. Letmε,t1,t2 be the minimum value of this minimization prob- lem.

In this part we show thatu_ε has exactly one layer near the interval [r₁, r₂].

Step 2.1. First we estimate the energy of transition layer. Let η >0 and θ > 0 be small numbers. Sinceuε →1 uniformly on [0, r1−θ] and uε → −1 uniformly on [r2 +θ,1−θ], we can find rε ∈ (r1 −θ, r2 +θ) such that uε(r) ≥ 1−η if r∈[0, rε],uε(r)<1−η forr−rε>0 small. Let ˜rε> rεbe such thatuε(r)≤ηif r∈[˜r_ε,1−θ],u_ε(r)> ηfor ˜r_ε−r >0 small. We may assume thatr_ε→r∈[r₁, r₂] and ˜r_ε→r˜∈[r₁, r₂]

We employ the so-called blow-up argument. Letvε(t) =uε(εt+rε). Then

−v_ε⁰⁰−εN−1 εt+rε

v⁰_ε=f(εt+rε, vε),

−1≤v_ε≤1 andv_ε(0) = 1−η. Sincer_ε→r∈[r₁, r₂], it is easy to see thatv_ε→v inC_loc¹ (R) and

−v⁰⁰=h(r)²(v−v³), t∈R.

andv(t)≥1−ηfort≤0. If we setv(t) =V(h(r)t), the functionV(t) satisfies

−V⁰⁰=V −V³ onR, V(0) = 1−η, V⁰(t)≥1−η t≤0.

(3.1) Hence by Lemma 3.3, the functionV is a unique solution for

−V⁰⁰=V −V³ onR, V(0) = 1−η, V⁰(t)<0 t≤0.

V(t)→ ±1 ast→ ∓∞.

(3.2)

Thus, we can find anR >0 large, such thatv(R) =η. Sincevε→vinC_loc¹ (R), we can find anRε∈(R−1, R+1), such thatv_ε⁰(r)<0 ifr∈[0, Rε] andvε(Rε) =−1+η.

Henceu⁰_ε(r)<0 ifr∈[rε, rε+εRε] anduε(rε+εRε) =−1 +η. Then we have Jε(uε, Br_ε+εR_ε(0)\Br_ε(0))

=ωN−1(r^N_ε⁻¹+oε(1))

Z r_ε+εR_ε r_ε

ε²

2 |u⁰_ε|²−F(t, uε)

dt

=ω_N−1(r^N_ε⁻¹+o_ε(1))ε Z Rε

0

1

2|v_ε⁰|²−F(εt+r_ε, v_ε)

dt

=ωN−1(r^N_ε⁻¹+oε(1))(β_h(r)+O(η) +oε(1))ε,

(3.3)

whereωN−1 is the area of the unit sphere inR^N, oε(1)→0 as ε→0,βh(s) is the positive value defined by

β_h(s)= Z +∞

−∞

1

2|w⁰_h(s)(t)|²+h(s)²(w²_h(s)−1)² 4

dt

=h(s) Z +∞

−∞

1

2|V⁰(t)|²+(V(t)²−1)²

4 dt

=h(s)β1

and w_h(s)(t) = V(h(s)t) for s ∈ [0,1]. We note that although the function V depends onη, the value

β₁= Z +∞

−∞

1

2|V⁰(t)|²+(V(t)²−1)²

4 dt

is independent ofη.

Step 2.2. We claimuεhas exactly one layer near the interval [r1, r2]. To showuε

has exactly one layer near the interval [r1, r2], it sufficient to prove the following claim

Claim. r˜_ε=r_ε+εR_ε.

Suppose that the claim is not true. Then we can find atε> rε+Rεεsuch that uε(r)<−1 +ηifr∈(rε+Rεε, tε),uε(tε) =−1 +η. Thus we can use the blow-up argument again at tε to deduce that there is a ˜tε = tε+εR˜ε with u⁰_ε(r) > 0 if r ∈ (tε,˜tε), uε(˜tε) = 1−η. We may assume that tε,˜tε → t as ε → 0 for some t∈[r₂, r₃]. Moreover

J_ε(u_ε, B˜t_ε(0)\B_tε(0)) =ω_N−1(t^N_ε⁻¹+o_ε(1))(β_h(t)+O(η))ε+o_ε(1) (3.4) Now we claim ˜t_ε ≥r₁. Suppose ˜t_ε< r₁. LetF_a(t) =Rt

−1(v−a)(1−v²)dv. Then for anyt >0 small ands∈[−1 +t,1−t],

Fa(1−t)−Fa(s)

=F0(1−t)−F0(s) +Fa(1−t)−F0(1−t)−Fa(s) +F0(s)

=(v²−1)² 4

1−t s −a

Z 1−t s

(1−v²)dv

(3.5)

Thus it follows from (3.5) that ifa <0, then

F_a(1−t)−F_a(s)>0 (3.6) fors∈[−1 +t,1−t]. Define

uε(r) :=

(1−η r∈[r_ε, r_ε+R_εε]∪[t_ε,˜t_ε],

−uε(r) r∈[rε+Rεε, tε].

By the assumption that ˜tε < r1 and using (3.6), we see F(r, uε) < F(r, uε) if r∈[rε,˜tε]. Hence, we obtain

Jε(uε, B˜tε(0)\Br_ε(0))< Jε(uε, Bt˜ε(0)\Br_ε(0)).

Thus we obtain a contradiction. Therefore we have that ˜tε≥r1.

Sincea(r)≥0 forr∈[r1,1], we see F(r, t)≤F(r,−1) = 0 if r∈[r1,1]. Since uε(r)∈(−1,−1 +η) forr∈[rε+Rεε, tε], we have

m_ε,rε,˜rε=J_ε(u_ε, B_rε+εRε(0)\Brε(0)) +J_ε(u_ε, B˜t_ε(0)\Btε(0)) +Jε(uε, Bt_ε(0)\Br_ε+εR_ε(0)) +Jε(uε, B˜r_ε(0)\Bt˜ε(0))

≥ω_N−1(r^N_ε⁻¹β_h(r)ε+t^N−1_ε β_h(t)ε) +O(ηε) +o(ε) + infn

− Z

B_tε(0)\B_rε+εRε(0)

F(r, w) :−1≤w≤1 +ηo + infn

− Z

B˜rε(0)\B˜tε(0)

F(r, w) :−1≤w≤1o

≥ω_N−1(r^N_ε⁻¹β_h(r)ε+t^N−1_ε β_h(t)ε) +O(ηε) +o(ε)

(3.7)

Now we give an upper bound formε,r_ε,˜r_ε. LetR >0 be such thatV(h(r)R) =η, whereV is a unique solution to (3.2). Defineuε by

uε(r) :=











V(h(r)^r−r_ε^ε) r∈[r_ε, r_ε+εR]

−1 +η−^η_ε(r−rε−εR) r∈[rε+εR, rε+εR+ε]

−1 r∈[r_ε+εR+ε,r˜_ε−ε]

−1 + ^η_ε(r−r˜ε+ε) r∈[˜rε−ε,r˜ε]

(3.8)

Now we note that|F(r, t)|=O(η) forr∈[rε,˜rε] and−1≤t≤ −1 +η . Then we have

m_ε,rε,˜rε ≤J_ε(u_ε, B_r˜ε(0)\B_rε(0))

≤Jε(uε, Brε+Rε(0)\Brε(0)) +Jε(uε, Br˜ε(0)\Br˜ε−ε(0)) +Jε(uε, Br˜_ε−ε(0)\Br_ε+εR(0))

≤ω_N−1r^N−1_ε (β_h(r)+O(η))ε+o(ε) +O(εη) +o(ε)

=ωN−1r^N−1_ε βh(r)+O(ηε) +o(ε)

(3.9)

By (3.7) and (3.9), we have

ω_N−1(r^N_ε⁻¹β_h(r)+t^N_ε⁻¹β_h(t))ε≤ω_N−1r^N−1_ε β_h(r)ε+O(εη) +o(ε) This is a contradiction. So we can conclude ˜r_ε=r_ε+εR_ε.

Part 3. It remains to prove that ifrεj →rfor some positive sequence {εj} con- verging to zero asj→ ∞thenrsatisfies

r^N−1h(r) = min

s∈[r1,r2]

s^N−1h(s).

Step 3.1. First we note that from Part 1, the functionuεsatisfies−1≤uε≤ −1+η forr∈[rε+εRε,1] in this case.

Step 3.2. Set H(s) = s^N−1h(s). Assume that the result is not true. Then there exists a subsequence of {rε} (denoted by r_ε) such that r_ε → r⁰ ∈ [r₁, r₂] and H(r⁰) > min_s∈[r1,r2]H(s). Then we can find a point t ∈ (r₁, r₂) such that H(r⁰)> H(t).

Now we give a lower estimate forJ_ε(u_ε). We have

Jε(uε) =Jε(uε, Br_ε(0)) +Jε(uε, Br_ε+εR_ε(0)\Br_ε(0)) +Jε(uε, B1(0)\Br_ε+R_εε(0)).

(3.10) First we note that 1−η ≤ uε(r) ≤1 for r≤rε and for sufficiently small η >0,

−F(r, u)≥ −F(r,1) (u∈[1−η,1]). We also remark that sincea(r)<0 forr < r1

and a(r) = 0 forr₁ ≤r ≤r₂ and a(r)>0 for r > r₂, we have −F(r,1) <0 for r < r₁ and −F(r,1) = 0 for r₁ ≤r≤r₂ and −F(r,1) >0 forr > r₂. Hence we have−Rrε

r₁ r^N−1F(r,1)dr≥0 and we obtain the estimate Jε(uε, Br_ε(0))≥ −

Z r_ε 0

r^N−1F(r, uε)dr

≥ − Z rε

0

r^N−1F(r,1)dr

=− Z r₁

0

r^N−1F(r,1)dr− Z r_ε

r₁

r^N−1F(r,1)dr

≥ − Z r₁

0

r^N−1F(r,1)dr=:A.

(3.11)

Using methods similar to those in the proof of (3.3), we obtain

J_ε(u_ε, B_rε+Rεε(0)\Brε(0))≥ω_N−1H(r⁰)β₁ε+O(ηε) +o(ε). (3.12)

Since −1 ≤ uε(r) ≤ −1 +η for r ≥ rε+εRε and for sufficiently small η > 0,

−F(r, u)≥ −F(r,−1) = 0 (u∈[−1,−1 +η]), we obtain the estimate Jε(uε, B1(0)\Br_ε+R_εε(0))≥ −

Z 1 rε+εRε

r^N−1F(r, uε)dr

≥ − Z 1

r_ε+εR_ε

r^N−1F(r,−1)dr= 0.

(3.13)

Thus we obtain

J(u_ε)≥A+ω_N−1H(r⁰)β₁ε+O(ηε) +o(ε). (3.14) Next we give an upper bound forJ_ε(u_ε). Consider the function

wε(r) :=



















1 r∈[0, t−ε]

1−^η_ε(r−t+ε) r∈[t−ε, t]

V h(t)^r−t_ε

r∈[t, t+εR⁰]

−1−^η_ε(r−t−εR⁰−ε) r∈[t+εR⁰, t+εR⁰+ε]

−1 r∈[t+εR⁰+ε,1], whereR⁰>0 is the number satisfyingV(h(t)R⁰) =−1 +η. Then

J_ε(u_ε)≤J_ε(w_ε)≤A+ω_N−1H(t)β₁ε+O(ηε) +o(ε). (3.15) By (3.14) and (3.15) we have a contradiction. The proof of Theorem 1.1 is complete.

The more complicate case, can be shown by a similar method (see Remark below).

Remark. We briefly show the more complicate case, that is, whenais the function as in Figure 2. More precisely we setI1:= [r1, r2] andI2:= [r3, r4] and we assume a >0 on [0, r1)∪(r4,1] anda <0 on (r3, r4).

r

r

2

r

3

r

4

a ( r )

u

I

I

ǭ

1

-1 O r

Figure 2. Special case of coefficienta(t)

Letη >0 andθ >0 be small numbers. As in Part 1, we can find pairs of numbers (r_1,ε, r_2,ε) and (R_1,ε, R_ε,2) satisfyingr_1,ε ∈(r₁−θ, r₂+θ), r_2,ε ∈(r₃−θ, r₄+θ),

sup_ε|R1,ε|<∞, sup_ε|R2,ε|<∞and

uε(r)<−1 +η for 0< r < r1,ε

u_ε(r_1,ε) =−1 +η uε(r1,ε+εR1,ε) = 1−η

uε(r)>1−η forr1,ε+εR1,ε< r < r2,ε

u_ε(r_2,ε) = 1−η uε(r2,ε+εR2,ε) =−1 +η uε(r)<−1 +η forr2,ε+εR2,ε< r <1

We assume thatr_1,εj →r₁ ∈I₁ and that r_2,εj →r₂∈I₂ for some sequence {ε_j} which converges to 0 as j→ ∞. In this case it is easy to show that the energy of global minimizerJ(uε) is estimated as follows

J_εj(u_εj)≥J_εj(u_εj, B_r2−ε(0)) +ε_jω_N−1H(r₂)β₁+B+O(ε_jη) +o(ε_j), (3.16) whereB=−Rr₃

r2 r^N−1F(r,1)dr.

Let us assume the result does not hold. ThenH(r1)>min_s∈I1H(s) orH(r2)>

min_s∈I2 hold. We assume H(r1) = min_s∈I1 and H(r2)> min_s∈I2H(s). We also assumer1=r1. We note that if H(r1)>mins∈I₁H(s) orr1∈intI1, the proof is more easy.

Let we take ˜r2 ∈intI2 such that H(r2)> H(˜r2) >mins∈I₂H(s) and consider the function

˜ uε(r) :=



























uε(r) on [0, r2−ε)

1 +^η_ε(r−r2) on [r2−ε, r2] 1 on [r2,r˜2−ε]

1−^η_ε(r−˜r2+ε) on [˜r2−ε,r˜2] V h(˜r2)^r−˜_ε^r2

on [˜r2,r˜2+εR⁰⁰]

−1−^η_ε(r−r˜2−εR⁰⁰−ε) on [˜r2+εR⁰⁰,r˜2+εR⁰⁰+ε]

−1 on [˜r2+εR⁰⁰+ε,1],

where V is the unique solution of (3.2) and R⁰⁰ is the unique value such that V(h(r₁)R⁰⁰) =−1 +η.

Sinceuεis global minimizer, we can estimate the energy ofJε(˜uε) as follows J_ε(u_ε)≤J_ε(˜u_ε)≤J_ε(u_ε, B_r2−ε(0)) +εω_N−1H(˜r₂)β₁+B+O(εη) +o(ε). (3.17) Then we have a contradiction from (3.16) and (3.17) by taking ε =εj and suffi- ciently largej.

Acknowledgments. The author would like to thank Professor Kazuhiro Kurata for his valuable advice and help, also to the anonymous referee for the numerous and useful comments.

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Hiroshi Matsuzawa

Numazu National College of Technology, Ooka 3600, Numazu-city, Shizuoka 410-8501, Japan

E-mail address:[email protected]